Compund Angle Identities and proof

1. Feb 20, 2014

FaraDazed

1. The problem statement, all variables and given/known data
prove using the compound angle identies, proove the following:
$$\frac{sin(A-B)}{cos(A)cos(B)}+\frac{sin(B-C)}{cos(B)cos(C)}+\frac{sin(C-A)}{cos(C)cos(A)}=0$$

2. Relevant equations
n/a

3. The attempt at a solution
I resolved it to
$$\frac{sin(A)cos(B)-cos(A)sin(B)}{cos(A)cos(B)}+\frac{sin(B)cos(C)-cos(B)sin(C)}{cos(B)cos(C)}+\frac{sin(C)cos(A)-cos(C)sin(A)}{cos(C)cos(A)}=0$$

And using wolfram alpha I found that the first term resolves to tan(A)-tan(B) and then I can see how it equals zero as all the tans would cancel out.

But I have no idea how each term simplifies to that.

Any help is really appreciated.

thanks.

2. Feb 20, 2014

CAF123

After you have expanded the numerators with the addition formulae (which you have done), you can simplify each fraction by taking out a factor.

3. Feb 20, 2014

Mentallic

In each fraction, you have something of the form

$$\frac{a+b}{c}$$

You should always check if splitting up the fraction into

$$\frac{a}{c}+\frac{b}{c}$$

could help you in any way. In this case, it does.

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